document.write( "Question 466310: (a) In a random sample of 150 persons having their lunch at the University cafeteria on meatless day it was observed that 20 percent preferred vegetable dishes.
\n" ); document.write( "• Find 95% confidence interval for P (proportion of those who preferred vegetables)
\n" ); document.write( "• How large a sample is needed, if we want to be 98% confident that our estimate of P is within 0.01\r
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Algebra.Com's Answer #319851 by stanbon(75887)\"\" \"About 
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(a) In a random sample of 150 persons having their lunch at the University cafeteria on meatless day it was observed that 20 percent preferred vegetable dishes.
\n" ); document.write( "• Find 95% confidence interval for P (proportion of those who preferred vegetables)
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\n" ); document.write( "p-hat = 0.2
\n" ); document.write( "ME = 1.96*sqrt[0.2*0.8/sqrt(150)] = 0.0640
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\n" ); document.write( "95% CI: 0.2-0.0640 < p < 0.2640
\n" ); document.write( "95% CI: 0.1360 < p < 0.2640
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\n" ); document.write( "• How large a sample is needed, if we want to be 98% confident that our estimate of P is within 0.01
\n" ); document.write( "---
\n" ); document.write( "n = [z/E]^2*pq
\n" ); document.write( "---
\n" ); document.write( "n = [2.32633/0.01]^2*0.2*0.8
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\n" ); document.write( "n = 8660 when rounded up
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\n" ); document.write( "cheers,
\n" ); document.write( "Stan H.
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